JEE MainChemistry

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Determining Unpaired Electrons in Octahedral Coordination Complexes

Q: Why does [ extCr(extCN)_6]^(3-) have 3 unpaired electrons even though extCN^(-) is a strong field ligand?

Because extCr^(3+) is a d^3 ion, and all 3 electrons singly occupy the lower-energy t_2g orbitals (t_2g^3) regardless of ligand field strength.

Match List I with List II:- [ List I List II; Coordination Complex Number of unpaired electrons; A. [Cr(CN)_6]^(3-) I. 0; B. [Fe(H_2O)_6]^(2+) II. 3; C. [Co(NH_3)_6]^(3+) III. 2; D. [Ni(NH_3)_6]^(2+) IV. 4; ] Choose the correct answer from the options given below:-

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AA-II, B-IV, C-I, D-IIIBA-IV, B-III, C-II, D-ICA-III, B-IV, C-I, D-IIDA-II, B-I, C-IV, D-III

Step-by-step solutionView

Correct answer

By determining the central metal ion's oxidation state, d-electron count, and the ligand field strength (weak vs. strong field), we find ${Cr}^{3 +}$ ( $d^{3}$ ) has 3, ${Fe}^{2 +}$ ( $d^{6}$ , weak field) has 4, ${Co}^{3 +}$ ( $d^{6}$ , strong field) has 0, and ${Ni}^{2 +}$ ( $d^{8}$ ) has 2 unpaired electrons, giving $A - II, B - IV, C - I, D - III$ .

Option analysis

Why each option works or fails

A · A-II, B-IV, C-I, D-III

The configuration and crystal field splitting for each metal complex are evaluated correctly. Correctly determined oxidation states, electronic configurations, and ligand splitting behavior for each complex.

B · A-IV, B-III, C-II, D-I

Treating the number of valence electrons from neutral atoms or misinterpreting d-counts without accounting for the formal oxidation states. First determine the oxidation state of the transition metal, remove electrons starting from the $4 s$ orbital, and then populate the split $d$ -orbitals according to ligand field strength.

C · A-III, B-IV, C-I, D-II

Assigning 2 unpaired electrons to $[e x t C r (e x t C N)_{6}]^{3 -}$ by mistakenly pairing its 3 electrons in the $t_{2 g}$ set under strong-field assumption. Remember that for $d^{1}$ , $d^{2}$ , and $d^{3}$ systems, Hund's rule applies regardless of ligand strength because the three $t_{2 g}$ orbitals each hold one unpaired electron before pairing can occur.

D · A-II, B-I, C-IV, D-III

Assuming that ligands behave identically across different metal ions, or reversing the strong/weak field pairing behavior for ${Fe}^{2 +}$ and ${Co}^{3 +}$ . Identify $H_{2} O$ as a weak-field ligand causing high-spin configuration ( $t_{2 g}^{4} e_{g}^{2}$ , 4 unpaired electrons) for ${Fe}^{2 +}$ , and ${NH}_{3}$ as a strong-field ligand causing low-spin configuration ( $t_{2 g}^{6} e_{g}^{0}$ , 0 unpaired electrons) for ${Co}^{3 +}$ .

Reviewed route

Solution

StepWorking

01Concept

Determine the oxidation state of the central metal ion, write its $3 d^{n}$ configuration, and apply Crystal Field Theory considering whether the ligand is strong-field (pairing occurs) or weak-field (high spin).

02Option verdict

A: ${Cr}^{3 +} (3 d^{3})$ has $t_{2 g}^{3} e_{g}^{0} ⟹ 3$ unpaired electrons (II). B: ${Fe}^{2 +} (3 d^{6})$ with weak field $H_{2} O$ has $t_{2 g}^{4} e_{g}^{2} ⟹ 4$ unpaired electrons (IV). C: ${Co}^{3 +} (3 d^{6})$ with strong field ${NH}_{3}$ has $t_{2 g}^{6} e_{g}^{0} ⟹ 0$ unpaired electrons (I). D: ${Ni}^{2 +} (3 d^{8})$ has $t_{2 g}^{6} e_{g}^{2} ⟹ 2$ unpaired electrons (III). Matching gives A-II, B-IV, C-I, D-III.

03Option verdict

Incorrectly matches A to IV and D to I, ignoring proper d-electron counting and CFT splitting.

04Option verdict

Matches A to III and D to II, miscounting ${Cr}^{3 +}$ as $d^{2}$ instead of $d^{3}$ .

05Option verdict

Swaps the electron configurations of ${Fe}^{2 +}$ (weak field) and ${Co}^{3 +}$ (strong field), assuming both pair up or both stay high-spin.

✓Discriminator

${Cr}^{3 +}$ is always $d^{3}$ (3 unpaired electrons $⟹ A-II$ ) and ${Co}^{3 +}$ with ${NH}_{3}$ is low-spin diamagnetic (0 unpaired electrons $⟹ C-I$ ), which uniquely identifies option A.

Hints that build this answer step by step

What are the oxidation states and $d$ -electron configurations of the central metal ions in $[e x t C r (e x t C N)_{6}]^{3 -}$ and $[e x t F e (e x t H_{2} e x t O)_{6}]^{2 +}$ ?
${Cr}^{3 +} (d^{3})$ and ${Fe}^{2 +} (d^{6})$
How many unpaired electrons do $[e x t C r (e x t C N)_{6}]^{3 -}$ and $[e x t F e (e x t H_{2} e x t O)_{6}]^{2 +}$ have under crystal field theory?
$[Cr (CN)_{6}]^{3 -}$ has 3 ( $t_{2 g}^{3}$ ), and $[Fe (H_{2} O)_{6}]^{2 +}$ has 4 ( $t_{2 g}^{4} e_{g}^{2}$ ).
What are the numbers of unpaired electrons for $[e x t C o (e x t {N H}_{3})_{6}]^{3 +}$ ( $d^{6}$ ) and $[e x t N i (e x t {N H}_{3})_{6}]^{2 +}$ ( $d^{8}$ )?
$[Co ({NH}_{3})_{6}]^{3 +}$ has 0 ( $t_{2 g}^{6}$ ), and $[Ni ({NH}_{3})_{6}]^{2 +}$ has 2 ( $t_{2 g}^{6} e_{g}^{2}$ ).

✓ Source and academic review

Question type: Match the following
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does

[e x t C r (e x t C N)_{6}]^{3 -}

have 3 unpaired electrons even though

e x t {C N}^{-}

is a strong field ligand?

Because $e x t {C r}^{3 +}$ is a $d^{3}$ ion, and all 3 electrons singly occupy the lower-energy $t_{2 g}$ orbitals ( $t_{2 g}^{3}$ ) regardless of ligand field strength.

Answer

By determining the central metal ion's oxidation state, d-electron count, and the ligand field strength (weak vs. strong field), we find ${Cr}^{3 +}$ ( $d^{3}$ ) has 3, ${Fe}^{2 +}$ ( $d^{6}$ , weak field) has 4, ${Co}^{3 +}$ ( $d^{6}$ , strong field) has 0, and ${Ni}^{2 +}$ ( $d^{8}$ ) has 2 unpaired electrons, giving $A - II, B - IV, C - I, D - III$ .

Why each option works or fails

A · correct: A-II, B-IV, C-I, D-III - The configuration and crystal field splitting for each metal complex are evaluated correctly. Correctly determined oxidation states, electronic configurations, and ligand splitting behavior for each complex.
B: A-IV, B-III, C-II, D-I - Treating the number of valence electrons from neutral atoms or misinterpreting d-counts without accounting for the formal oxidation states. First determine the oxidation state of the transition metal, remove electrons starting from the $4 s$ orbital, and then populate the split $d$ -orbitals according to ligand field strength.
C: A-III, B-IV, C-I, D-II - Assigning 2 unpaired electrons to $[e x t C r (e x t C N)_{6}]^{3 -}$ by mistakenly pairing its 3 electrons in the $t_{2 g}$ set under strong-field assumption. Remember that for $d^{1}$ , $d^{2}$ , and $d^{3}$ systems, Hund's rule applies regardless of ligand strength because the three $t_{2 g}$ orbitals each hold one unpaired electron before pairing can occur.
D: A-II, B-I, C-IV, D-III - Assuming that ligands behave identically across different metal ions, or reversing the strong/weak field pairing behavior for ${Fe}^{2 +}$ and ${Co}^{3 +}$ . Identify $H_{2} O$ as a weak-field ligand causing high-spin configuration ( $t_{2 g}^{4} e_{g}^{2}$ , 4 unpaired electrons) for ${Fe}^{2 +}$ , and ${NH}_{3}$ as a strong-field ligand causing low-spin configuration ( $t_{2 g}^{6} e_{g}^{0}$ , 0 unpaired electrons) for ${Co}^{3 +}$ .

Step-by-step solution

Concept: Determine the oxidation state of the central metal ion, write its $3 d^{n}$ configuration, and apply Crystal Field Theory considering whether the ligand is strong-field (pairing occurs) or weak-field (high spin).
Option verdict: A: ${Cr}^{3 +} (3 d^{3})$ has $t_{2 g}^{3} e_{g}^{0} ⟹ 3$ unpaired electrons (II). B: ${Fe}^{2 +} (3 d^{6})$ with weak field $H_{2} O$ has $t_{2 g}^{4} e_{g}^{2} ⟹ 4$ unpaired electrons (IV). C: ${Co}^{3 +} (3 d^{6})$ with strong field ${NH}_{3}$ has $t_{2 g}^{6} e_{g}^{0} ⟹ 0$ unpaired electrons (I). D: ${Ni}^{2 +} (3 d^{8})$ has $t_{2 g}^{6} e_{g}^{2} ⟹ 2$ unpaired electrons (III). Matching gives A-II, B-IV, C-I, D-III.
Option verdict: Incorrectly matches A to IV and D to I, ignoring proper d-electron counting and CFT splitting.
Option verdict: Matches A to III and D to II, miscounting ${Cr}^{3 +}$ as $d^{2}$ instead of $d^{3}$ .
Option verdict: Swaps the electron configurations of ${Fe}^{2 +}$ (weak field) and ${Co}^{3 +}$ (strong field), assuming both pair up or both stay high-spin.
Discriminator: $t_{2 g}^{3} e_{g}^{0} ⟹ 3$ is always $d^{3}$ (3 unpaired electrons $⟹ A-II$ ) and ${Co}^{3 +}$ with ${NH}_{3}$ is low-spin diamagnetic (0 unpaired electrons $⟹ C-I$ ), which uniquely identifies option A.

Determining Unpaired Electrons in Octahedral Coordination Complexes

What feels right?

Why each option works or fails

Solution

01Concept

02Option verdict

03Option verdict

04Option verdict

05Option verdict

✓Discriminator

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